2. Impulse and Momentum

3. ASTRONAUT Edward H. White II floats in the zero gravity of space. By firing the gas-powered gun, he gains momentum and maneuverability. Credit: NASA

4. IMPULSE tt F J = F  t Impulse: Impulse J is a force F acting for a small time interval  t.

5. Example 1: The face of a golf club exerts an average force of 4000 N for 0.002 s. What is the impulse imparted to the ball? tt F J = F  t Impulse: J = (4000 N)(0.002 s) J = 8.00 N  s The unit for impulse is the Newton-second (N s)

6. Impulse from a Varying Force Normally, a force acting for a short interval is not constant. It may be large initially and then play off to zero as shown in the graph. F time, t In the absence of calculus, we use the average force F avg.

7. Example 2: Two flexible balls collide. The ball B exerts an average force of 1200 N on ball A. How long were the balls in contact if the impulse is 5 N s?  t = 0.00420 s The impulse is negative; the force on ball A is to the left. Unless told otherwise, treat forces as average forces. BA

8. Impulse Changes Velocity Consider a mallet hitting a ball: F Impulse = Change in “mv”

9. Momentum Defined Momentum p is defined as the product of mass and velocity, mv. Units: kg m/s p = mv Momentum m = 1000 kg v = 16 m/s p = (1000 kg)(16 m/s) p = 16,000 kg m/s

10. Impulse and Momentum Impulse = Change in momentum F  t = mv f - mv o tt F mv A force F acting on a ball for a time  t increases its momentum mv.

11. Example 3: A 50-g golf ball leaves the face of the club at 20 m/s. If the club is in contact for 0.002 s, what average force acted on the ball? tt F mv Given: Given: m = 0.05 kg; v o = 0;  t = 0.002 s; v f = 20 m/s + Choose right as positive. F  t = mv f - mv o F (0.002 s) = (0.05 kg)(20 m/s) Average Force: F = 500 N 0

12. Vector Nature of Momentum Consider the change in momentum of a ball that is dropped onto a rigid plate: vovo vfvf A 2-kg ball strikes the plate with a speed of 20 m/s and rebounds with a speed of 15 m/s. What is the change in momentum? +  p = mv f - mv o = (2 kg)(15 m/s) - (2 kg)(-20 m/s)  p = 30 kg m/s + 40 kg m/s  p = 70 kg m/s

13. Directions Are Essential 1. Choose and label a positive direction. + vfvf v0v0 v f – v 0 = (10 m/s) – (-30 m/s) 2. A velocity is positive when with this direction and negative when against it. Assume v 0 is 30 m/s to the left and v f is 10 m/s to the right. What is the change in velocity  v? v f = +10 m/s v 0 = -30 m/s

14. Example 4: A 500-g baseball moves to the left at 20 m/s striking a bat. The bat is in contact with the ball for 0.002 s, and it leaves in the opposite direction at 40 m/s. What was average force on ball? 40 m/s tt F 20 m/s m = 0.5 kg + - + F  t = mv f - mv o F(0.002 s) = (0.5 kg)(40 m/s) - (0.5 kg)(-20 m/s) v o = -20 m/s; v f = 40 m/s Continued...

15. Example Continued: 40 m/s tt F 20 m/s m = 0.5 kg + - + F  t = mv f - mv o F(0.002 s) = (0.5 kg)(40 m/s) - (0.5 kg)(-20 m/s) F(0.002 s) = (20 kg m/s) + (10 kg m/s) F(0.002 s) = 30 kg m/s F = 15,000 N

16. Impulse in Two Dimensions F x  t = mv fx - mv ox + vovo F FxFx FyFy vfvf v fx v fy A baseball with an initial velocity v o hits a bat and leaves with v f at an angle. Horizontal and vertical impulse are independent. F y  t = mv fy - mv oy F = F x i + F y jv o = v ox i + v oy jv f = v x i + v y j +

17. Example 5: A 500-g baseball travelling at 20 m/s leaves a bat with a velocity of 50 m/s at an angle of 30 0. If  t = 0.002 s, what was the average force F? + vovo F FxFx FyFy vfvf v fx v fy + 30 0 -20 m/s 50 m/s v ox = -20 m/s; v oy = 0 v fx = 50 Cos 30 0 = 43.3 m/s v fy = 50 Sin 30 0 = 25 m/s First consider horizontal: F x  t = mv fx - mv ox F x (.002 s) = (0.5 kg)(43.3 m/s) - (0.5 kg)(-20 m/s)

18. Example Continued... F x (.002 s) = (0.5 kg)(43.3 m/s) - (0.5 kg)(-20 m/s) + vovo F FxFx FyFy vfvf v fx v fy + 30 0 20 m/s 50 m/s F x (.002 s) = 21.7 kg m/s + 10 kg m/s) F x = 15.8 kN Now apply to vertical: F y  t = mv fy - mv oy 0 F y (.002 s) = (0.5 kg)(25 m/s) F y = 6.25 kN F = 17.0 kN, 21.5 0 and

19. Summary of Formulas: Momentum p = mv Impulse I= F avg  t Impulse = Change in momentum F  t = mv f - mv o